22 Aug
2009
22 Aug
'09
7 p.m.
Dear all,
let A and B be sets and R,S \inc A x B. Let A_0 be the subset of A consisting of all a such that (a,b) \in R iff (a,b) \in S. Then it seems to me that the inclusion function of A_0 into A is the equalizer of R and S.
I already answered Sergei privately, but this should be corrected publicly: Rel does not have equalisers. In fact, it has almost no equalisers. For a counterexample, consider the sets A = {0,1} and B = {0}, and the parallel relations R = A x B and S = {(0,0)}. Their equaliser, if it existed, must be contained in T = {(0,0)}. But T' = {0} x A also satisfies RT'=ST', but does not factor through any subrelation of T. Best wishes, Chris Heunen [For admin and other information see: http://www.mta.ca/~cat-dist/ ]